Find the basis and dimension for the column space of the matrix \(A = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix}\)

Step 1 ：Step 1: To find the basis for the column space of the matrix, we first need to find the column vectors of the matrix. The column vectors of the matrix are \(\begin{bmatrix} 1 \ 4 \ 7 \end{bmatrix}\), \(\begin{bmatrix} 2 \ 5 \ 8 \end{bmatrix}\), and \(\begin{bmatrix} 3 \ 6 \ 9 \end{bmatrix}\)

Step 2 ：Step 2: Next, we need to find the linearly independent column vectors. We can accomplish this by reducing the matrix to its row echelon form, or RREF. The RREF of the matrix is \(\begin{bmatrix} 1 & 0 & -1 \ 0 & 1 & 2 \ 0 & 0 & 0 \end{bmatrix}\)

Step 3 ：Step 3: The columns of the RREF matrix that have leading ones (pivot columns) correspond to the linearly independent columns of the original matrix. Here, the pivot columns are the first and second columns. Therefore, the linearly independent column vectors of the original matrix are \(\begin{bmatrix} 1 \ 4 \ 7 \end{bmatrix}\) and \(\begin{bmatrix} 2 \ 5 \ 8 \end{bmatrix}\)

Step 4 ：Step 4: So, the basis for the column space of the matrix is the set of linearly independent column vectors, which is \(\{\begin{bmatrix} 1 \ 4 \ 7 \end{bmatrix}, \begin{bmatrix} 2 \ 5 \ 8 \end{bmatrix}\}\)

Step 5 ：Step 5: The dimension of the column space is the number of linearly independent column vectors, which is 2.